OFFSET
1,2
COMMENTS
Suggested by the following question from Andreas Weingartner. Prove or disprove: there exists an epsilon>0 such that no natural number has the property that in base 2 as well as in base 3 at most (epsilon)*100% of the digits are nonzero.
LINKS
Robert Harley and Benne de Weger, Table of n, a(n) for n = 1..23 (terms 1 through 21 were computed by Robert Harley; terms 22 and 23 by Benne de Weger, Oct 16 2014)
FORMULA
Heuristically one might expect (except possibly for some small examples) such an eps for bases a and b at the solution of: (log(a-1)*log(b)+log(b-1)*log(a))*eps - log(a)*log(b) = (eps*log(eps)+(1-eps)*log(1-eps))*(log(a)+log(b)). For bases 2 and 3 this is about 0.131737... (Formula and value for eps corrected by Benne de Weger, Oct 13 2014)
EXAMPLE
18 = 10010 in base 2, f(18) = 2/5, 18 = 200 in base 3, g(18) = 1/3, h(n) = max {2/5, 1/3} = 2/5 and this is a new (low) record.
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Robert Harley, Dec 28 2002
STATUS
approved